Q.

Let A be the set of first 101 terms of an A.P., whose first term is 1 and the common difference is 5 and let B be the set of first 71 terms of an A.P., whose first term is 9 and the common difference is 7. Then the number of elements in A∩B, which are divisible by 3, is:          [2026]

1 4  
2 5  
3 6  
4 7  

Ans.

(2)

A={1,6,11,16,...,501}⏟101terms

⇒ Tn=5n–4, where 1≤n≤101

B={9,16,23,...,499}⏟71terms

⇒ Tm=7m+2, where 1≤m≤71

For common difference of

A∩B=LCM {5,7}=35

A∩B={16,51,...}

⇒ TK=35K–19

⇒ 16≤35K–19≤499

⇒ 35≤35K≤518

⇒ 1≤K≤14.8

∴  K has 14 values

Now, terms divided by 3 in A∩B are {51, 156, 261, 366, 471}                [∴  LCM of 35 and 3 is 105]

Hence, the required number of terms is 5.